已知数列《an>的前n项和为sn,a1=2,na=sn,求s2011
问题描述：
已知数列《an>的前n项和为sn,a1=2,na
答
由na(n+1)=(n+2)Sn,得a(n+1)=((n+2)/n) Sn∵a(n+1)=S(n+1)-Sn∴S(n+1)-Sn=((n+2)/n) SnS(n+1)=((n+2)/n) Sn+ Sn=(2(n+1)/n) SnS(n+1)/(n+1)= 2 Sn/n设bn=Sn/n,则b1=S1=a1=2b(n+1)=2bnb(n+1)/bn=2∴{bn}是首项为2、公...